Problem: Find $\int \dfrac{1}{6x^2+36x+78}\,dx$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{12} \text{arctan}\left(\dfrac{x+3}{2}\right)+C$ (Choice B) B $\dfrac{1}{12} \text{arcsin}\left(\dfrac{x+3}{2}\right)+C$ (Choice C) C $\dfrac12\text{arcsin}\left(\dfrac{x+3}{2}\right)+C$ (Choice D) D $\dfrac12\text{arctan}\left(\dfrac{x+3}{2}\right)+C$
Explanation: The integrand is in the form $\dfrac{1}{p(x)}$ where $p(x)$ is a quadratic expression. This suggests that we should rewrite $p(x)$ by completing the square. Specifically, we will rewrite $p(x)$ as a constant multiple of $(x+h)^2+k^2$. Then, we will be able to integrate using the following formula, which is based on the derivative of the inverse tangent function: $\int \dfrac{1}{(x+ h)^2+ k^2}\,dx=\dfrac{1}{ k} \text{arctan}\left(\dfrac{x+ h}{ k}\right)+C$ [Why is this formula true?] We start by rewriting $p(x)$ as a constant multiple of $(x+ h)^2+ k^2$ : $\begin{aligned} 6x^2+36x+78&=6[x^2+6x+13] \\\\ &=6[x^2+6x+9+4] \\\\ &=6[(x+3)^2+4] \\\\ &=6[(x+{3})^2+{2}^2] \end{aligned}$ Now we can find the integral: $\begin{aligned} &\phantom{=}\int \dfrac{1}{6x^2+36x+78}\,dx \\\\ &=\int\dfrac{1}{6[(x+3)^2+2^2]}\,dx \\\\ &=\dfrac16\int\dfrac{1}{(x+{3})^2+{2}^2}\,dx \\\\ &=\dfrac16\cdot\dfrac{1}{{2}} \text{arctan}\left(\dfrac{x+{3}}{{2}}\right)+C \\\\ &=\dfrac{1}{12} \text{arctan}\left(\dfrac{x+3}{2}\right)+C \end{aligned}$ In conclusion, $\int \dfrac{1}{6x^2+36x+78}\,dx=\dfrac{1}{12} \text{arctan}\left(\dfrac{x+3}{2}\right)+C$